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ISEE Middle Level Quantitative : Numbers and Operations

Study concepts, example questions & explanations for ISEE Middle Level Quantitative

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All ISEE Middle Level Quantitative Resources

6 Diagnostic Tests 110 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}3 3\\ \times\ 23\end{array}}{\ \ \ \ }$

Possible Answers:

554

627

834

759

960

Correct answer:

759

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}3{\color{Red} 3}\\ \times\ {2\color{Red} 3}\end{array}}{\ \ \ \ \ 9}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3}3\\ \times\ {2\color{Red} 3}\end{array}}{\ \ \ 99}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a as a place holder in the ones position.

$\frac{\begin{array}[b]{r}3 3\\ \times\ 23\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}3 {\color{Red} 3}\\ \times\ {\color{Red} 2}3\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ \ 60}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3} {\color{Black} 3}\\ \times\ {\color{Red} 2}3\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ 660}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}33\\ \times \ \ \ 23\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 99}\\ \ +\ {\color{Red} 660} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 759}$

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Example Question #2 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}14\\ \times\ 22\end{array}}{\ \ \ \ }$

Possible Answers:

432

226

396

308

476

Correct answer:

308

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}1{\color{Red} 4}\\ \times\ {2\color{Red} 2}\end{array}}{\ \ \ \ \ 8}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1}4\\ \times\ {2\color{Red} 2}\end{array}}{\ \ \ 28}$

$\frac{\begin{array}[b]{r}1 4\\ \times\ 22\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}1 {\color{Red} 4}\\ \times\ {\color{Red} 2}2\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ \ 80}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1} {\color{Black} 4}\\ \times\ {\color{Red} 2}2\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ 280}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}14\\ \times \ \ \ 22\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 28}\\ \ +\ {\color{Red} 280} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 308}$

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Example Question #2 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}43\\ \times\ 35\end{array}}{\ \ \ \ }$

Possible Answers:

876

804

985

$1\textup,505$

932

Correct answer:

$1\textup,505$

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}^14{\color{Red} 3}\\ \times\ {3\color{Red} 5}\end{array}}{\ \ \ \ \ 5}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 1}{\color{Red} 4}3\\ \times\ {3\color{Red} 5}\end{array}}{\ \ \ 215}$

$\frac{\begin{array}[b]{r}43\\ \times\ 35\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}4 {\color{Red} 3}\\ \times\ {\color{Red} 3}5\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ \ \ \ 90}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 3}\\ \times\ {\color{Red} 3}5\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ 1290}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}43\\ \times \ \ \ 35\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 215}\\ \ +\ {\color{Red} 1290} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 1505}$

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Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}40\\ \times\ 10\end{array}}{\ \ \ \ }$

Possible Answers:

410

444

400

440

420

Correct answer:

400

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}4{\color{Red} 0}\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ \ \ 0}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red}4}0\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ 00}$

$\frac{\begin{array}[b]{r}40\\ \times\ 10\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}4 {\color{Red} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ 00}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ 400}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}40\\ \times \ \ \ 10\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 00}\\ \ +\ {\color{Red} 400} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 400}$

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Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}52\\ \times\ 43\end{array}}{\ \ \ \ }$

Possible Answers:

$2\textup,136$

$2\textup,236$

$1\textup,824$

$2\textup,080$

$1\textup,328$

Correct answer:

$2\textup,236$

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}5{\color{Red} 2}\\ \times\ {4\color{Red} 3}\end{array}}{\ \ \ \ \ 6}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5}2\\ \times\ {4\color{Red} 3}\end{array}}{\ \ 156}$

$\frac{\begin{array}[b]{r}52\\ \times\ 43\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}5 {\color{Red} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ 80}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ 2080}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}52\\ \times \ \ \ 43\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 156}\\ \ +\ {\color{Red} 2080} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 2236}$

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Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}56\\ \times\ 17\end{array}}{\ \ \ \ }$

Possible Answers:

874

781

752

808

952

Correct answer:

952

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}^45{\color{Red} 6}\\ \times\ {1\color{Red} 7}\end{array}}{\ \ \ \ \ 2}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 4}{\color{Red} 5}6\\ \times\ {1\color{Red} 7}\end{array}}{\ \ 392}$

$\frac{\begin{array}[b]{r}5 6\\ \times\ 17\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}5 {\color{Red} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ 60}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ 560}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}56\\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 392}\\ \ +\ {\color{Red} 560} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 952}$

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Example Question #31 : Whole Numbers

Solve:

$\frac{\begin{array}[b]{r} 47\\ \times\ 12\end{array}}{\ \ \ \ }$

Possible Answers:

673

704

622

564

582

Correct answer:

564

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}^14{\color{Red} 7}\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ \ 4}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 1}{\color{Red} 4}7\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ 94}$

$\frac{\begin{array}[b]{r}47\\ \times\ 12\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}4 {\color{Red} 7}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ \ 70}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 7}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ 470}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}47\\ \times \ \ \ 12\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 94}\\ \ +\ {\color{Red} 470} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 564}$

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Example Question #12 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}15\\ \times\ 11\end{array}}{\ \ \ \ }$

Possible Answers:

175

150

160

170

165

Correct answer:

165

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}1{\color{Red} 5}\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ \ \ 5}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1}5\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ 15}$

$\frac{\begin{array}[b]{r}15\\ \times\ 11\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}1 {\color{Red} 5}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ \ 50}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1} {\color{Black} 5}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ 150}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}15\\ \times \ \ \ 11\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 15}\\ \ +\ {\color{Red} 150} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 165}$

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Example Question #13 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}30\\ \times\ 20\end{array}}{\ \ \ \ }$

Possible Answers:

300

500

200

400

600

Correct answer:

600

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}3{\color{Red} 0}\\ \times\ {2\color{Red} 0}\end{array}}{\ \ \ \ \ 0}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3}0\\ \times\ {2\color{Red} 0}\end{array}}{\ \ \ 00}$

$\frac{\begin{array}[b]{r}30\\ \times\ 20\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}3 {\color{Red} 0}\\ \times\ {\color{Red} 2}0\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ 00}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3} {\color{Black} 0}\\ \times\ {\color{Red} 2}0\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ 600}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}30\\ \times \ \ \ 20\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 00}\\ \ +\ {\color{Red} 600} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 600}$

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Example Question #1001 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times\phantom{0}3\space{\,}5 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1785

17850

1775

1795

Correct answer:

1785

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 35 is the multiplier and 51 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 5 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,} \,5\end{array}$

Then, we multiply 5 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,}2\,5\,5\end{array}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,}2\,5\,5 \\ \, 0 \end{array}$

Next, we multiply 3 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,}2\,5\,5 \\ \, 3 \, 0\end{array}$

Then, we multiply 3 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,}2\,5\,5 \\ \, 1\, 5\, 3 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}1\\ \times \phantom{0} 3\space{\,}5 \\ \hline \phantom{\,}2\,5\,5 \\+\, 1\, 5\, 3 \, 0\\ \hline 1\, 7 \, 8\, 5\end{array}$

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All ISEE Middle Level Quantitative Resources

6 Diagnostic Tests 110 Practice Tests Question of the Day Flashcards Learn by Concept

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ISEE Middle Level Quantitative : Numbers and Operations

Study concepts, example questions & explanations for ISEE Middle Level Quantitative

All ISEE Middle Level Quantitative Resources

Example Questions

Example Question #4 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #2 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #2 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #31 : Whole Numbers

Example Question #12 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #13 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1001 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

All ISEE Middle Level Quantitative Resources

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